Question: Let $g(x)=-2x^4+8x^3-12x^2$. For what values of $x$ does the graph of $g$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-8$ (Choice B) B $x=-6$ (Choice C) C $x=1$ (Choice D) D $g$ has no points of inflection.
We can find the inflection points of the graph of $g$ by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=-24(x-1)^2$. $g''(x)=0$ for $x=1$. Since $g''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection point is $x=1$. Our possible inflection points divide the number line into two intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $x<1$ $x>1$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $x<1$ $x=0$ $g''(0)=-24<0$ $g$ is concave down $\cap$ $x>1$ $x=2$ $g''(2)=-24<0$ $g$ is concave down $\cap$ We can see that the graph of $g$ does not change concavity at $x=1$. Therefore, $x=1$ is not a point of inflection. In conclusion, $g$ has no points of inflection.